Class 12 RD Sharma Solutions - Chapter 6 Determinants Exercise Ex. 6.6 | Set 3

Question 38. Write the value of the determinant .

Solution:

 We have,

A = 

|A| = 

On taking 2x common from R₃ we get,

|A| = 

As R₁ and R₃ are identical we get

|A| = 0

Therefore, the value of the determinant is 0.

Question 39. If |A| = 2, where A is 2 × 2 matrix, find |adj A|.

Solution:

Given that |A| = 2 and the order of A matrix is 2 x 2

As we know that |adj A| = |A|^n-1

Here the value of n is 2

So, 

|adj A| = |2|^2-1

= 2

Therefore, the value of the |adj A| is 2.

Question 40. What is the value of the determinant ?

Solution:

We have,

A = 

|A| = 

= 0(18 - 20) - 2(12 - 16) + 0(10 - 12) 

= 0 + 8 + 0

= 8

Therefore, the value of the determinant is 8.

Question 41. For what value of x is the matrix  singular?

Solution:

As we know, a matrix is singular matrix, when the value of its determinant is 0.

Given that,

A = 

So, 

|A| = 

=> 

=> (6 - x) - 4(3 - x) = 0

=> 6 - x - 12 + 4x = 0

=> 3x - 6 = 0  

=> 3x = 6

=> x = 6/3

=> x = 2

Therefore, the value of x is 2.

Question 42. A matrix A of order 3 × 3 is such that |A| = 4. Find the value of |2 A|.

Solution:

We have,

A matrix A is of order 3 × 3 so the value of n is 3.

And |A| = 4.

As we know,

=> |K A| = Kⁿ |A|

So, 

|2 A| = 2³ (4)

= 8 (4)

= 32

Therefore, the value of |2 A| is 32.

Question 43. Evaluate: .

Solution:

We have,

A = 

|A| = 

= cos 15° cos 75° - sin 15° sin 75°

As cos A cos B - sin A sin B = cos (A + B), we get 

= cos (15° + 75°) 

= cos 90°

= 0

Question 44. If A = , write the cofactor of the element a₃₂.

Solution:

We have, 

A = 

So, the minor of a₃₂ is,

M₃₂ = 

= 5 - 16 

= -11  

Now, the cofactor of a₃₂ is,

A₃₂ = (−1)³⁺² M₃₂ 

= 11

Therefore, the cofactor of element a₃₂ is 11.

Question 45. If , then write the value of x.

Solution:

We have,

On expanding the determinants of both sides, we get

=> (x + 1) (x + 2) - (x - 1) (x - 3) = 12 + 1

=> x² + 3x + 2 - x² + 4x - 3 = 13

=> 7x - 1 = 13

=> 7x = 14

=> x = 2 

Therefore, the value of x is 2.

Question 46. If , then write the value of x.

Solution:

We are given,

=> 

On expanding the determinants of both sides, we get

=> (2x) (x + 1) - 2 (x + 1) (x + 3) = 3 - 15  

=> (x + 1) (2x - 2x - 6) = -12 

=> -6x - 6 = - 12  

=> -6x = -6

=> x = 1

Therefore, the value of x is 1. 

Question 47. If , find the value of x.

Solution:

We have,

=> 

On expanding the determinants of both sides, we get

=> 12x + 14 = 32 - 42  

=> 12x + 14 = -10

=> 12x = -24

=> x = -24/12  

=> x = -2

Therefore, the value of x is -2. 

Question 48. If , write the value of x.

Solution:

Here, we have,

=>  

On expanding the determinants of both sides, we get

=> 2x² - 40 = 18 + 14  

=> 2x² - 40 = 32

=> 2x² = 72  

=> x² = 72/2

=> x² = 36

=> x = ±6

Therefore, the value of x is ±6.

Question 49. If A is a 3 × 3 matrix, |A| ≠ 0 and |3A| = k |A| then write the value of k.

Solution:

We are given,

A is a 3 × 3 matrix.

Also |A| ≠ 0 and |3A| = k |A|.

Let us considered A =   

3A = 

|3A| = 

On taking 3 common from each row we get,

= 

= 27 |A|

Therefore, the value of k is 27. 

Question 50. Write the value of the determinant .

Solution:

We have,

A = 

|A| = 

On expanding the determinant we have,

= p² - (p + 1) (p - 1)

= p² - (p² - 1)

= p² - p² + 1

= 1

Therefore, the value of the determinant is 1.

Question 51. Write the value of the determinant .

Solution:

We have,

A = 

|A| = 

On applying R₁ -> R₁ + R₂ we get,

= 

On taking x + y + z common from R₁ we have,

=  

On applying R₃ -> R₃ + 3 R₁ we get,

=  

= 

On expanding along the last row we get,

= 0

Therefore, the value of the determinant is 0.

Question 52. If A = , then for any natural number, find the value of Det(Aⁿ).

Solution:

Given that, A = 

=> A² = 

= 

= 

Similarly, Aⁿ = 

So,  

|Aⁿ| = 

= (cos nθ) (cos nθ) + (sin nθ) (sin nθ)

= cos² (nθ) + sin² (nθ)

= 1

Therefore, Det(Aⁿ) = 1.

Question 53. Find the maximum value of .

Solution:

We have,

A = 

|A| = 

On applying R₂ -> R₂ - R₁ and R₃ -> R₃ - R₁, we get

|A| =   

= sin θ cos θ

= (sin 2θ)/2

We know that −1 ≤ sin2θ ≤ 1.

So, the maximum value of |A| = (1/2) (1) 

= 1/2

Therefore, the maximum value is 1/2.

Question 54. If x ∈ N and  = 8, then find the value of x.

Solution:

Here we have,

A = 

|A| = 8

On expansion we get,

=> (x + 3) (2x) - (-2) (-3x) = 8

=> 2 x² + 6x - 6x = 8

=> 2 x² = 8

=> 2x² - 8 = 0

=> x² - 4 = 0

=> x² = 4

As x ∈ N, we get

=> x = 2 

Therefore, the value of x is 2.

Question 55. If , write the value of x.

Solution:

We have,

A = 

|A| = 

=> 

On expanding along R₁, we get

=> x (-x² - 1) - sin θ (-x sin θ - cos θ) + cos θ (-sin θ + x cos θ) = 8

=> -x³ - x + x sin² θ + sin θ cos θ - sin θ cos θ + x cos² θ = 8

=> -x³ - x + x (sin² θ + cos² θ) = 8

=> -x³ - x + x = 8

=> x³ + 8 = 0

=> x = -2

Therefore, the value of x is -2.

Question 56. If A is a 3 × 3 invertible matrix, then what will be the value of k if det(A^–1) = (det A)^k.

Solution:

Given that A is a 3 × 3 invertible matrix.

So, we know that

Therefore we get,

=> 

As we know that, 

= 

= 

= |A|

|A^-1| = |A^k|, we get

=> k = 1

Therefore, the value of k is 1.